an asteroid has a diameter of 10km and is at a distance of 10^6km, what angle in degrees does it subtend?
+26388 An asteroid has a diameter of 10km and is at a distance of 10^6km, what angle in degrees does it subtend ?
$$\tan{ ( \frac{ \Theta\ensurement{^{\circ}} } {2} ) } = \dfrac{ 5\ km } { 10^6\ km } = 5*10^{-7}\\\\
\frac{ \Theta\ensurement{^{\circ}} } {2} = 0.00002864789\ensurement{^{\circ}}\\\\
\Theta\ensurement{^{\circ}} =0.00005729578\ensurement{^{\circ}}\\\\
\Theta^{'}=\Theta\ensurement{^{\circ}} \times
\dfrac{60^{'}}
{ 1\ensurement{^{\circ}} }
= (0.00005729578 * 60)^{'} = 0.00343774677^{'}\\\\
\Theta^{"}=\Theta^{'}\times
\dfrac{ 60^{"} }
{ 1^{'} }
= (0.00343774677* 60)^{"} = 0.20626480625^{"}$$
+26388 An asteroid has a diameter of 10km and is at a distance of 10^6km, what angle in degrees does it subtend ?
$$\tan{ ( \frac{ \Theta\ensurement{^{\circ}} } {2} ) } = \dfrac{ 5\ km } { 10^6\ km } = 5*10^{-7}\\\\
\frac{ \Theta\ensurement{^{\circ}} } {2} = 0.00002864789\ensurement{^{\circ}}\\\\
\Theta\ensurement{^{\circ}} =0.00005729578\ensurement{^{\circ}}\\\\
\Theta^{'}=\Theta\ensurement{^{\circ}} \times
\dfrac{60^{'}}
{ 1\ensurement{^{\circ}} }
= (0.00005729578 * 60)^{'} = 0.00343774677^{'}\\\\
\Theta^{"}=\Theta^{'}\times
\dfrac{ 60^{"} }
{ 1^{'} }
= (0.00343774677* 60)^{"} = 0.20626480625^{"}$$
